Talk:Homology braid II (Ex)

Revision as of 08:58, 29 March 2012

1) The diagram is composed of the exact sequences of the pair $<wikitex>; '''1)''' The diagram is composed of the exact sequences of the pair $(\widetilde W,\widetilde M)$, and of the triple $(\widetilde W,\widetilde M\cup\widetilde M',\widetilde M)$, and of the analogous sequences with $\widetilde M$ and $\widetilde M'$ swapped. Because $\widetilde M$ and $\widetilde M'$ are disjoint, we have an isomorphism $H_k(\widetilde M\cup\widetilde M',\widetilde M')\cong H_k(\widetilde M)$ and another one with $\widetilde M$ and $\widetilde M'$ swapped. Both are implicit in the braid. The squares that are half visible on the left and right hand side obviously commute. The upper triangles commute because of the following diagram, $$\xymatrix{ H_{i+1}(\widetilde W,\widetilde M) \ar[d] \ar[r]^{\partial} & H_i(\widetilde M) \ar[d]^{\cong} \ar[r] & H_i(\widetilde W) \ar[d] \ H_{i+1}(\widetilde W,\widetilde M\cup\widetilde M') \ar[r]^{\partial} & H_i(\widetilde M\cup\widetilde M',\widetilde M') \ar[r] & H_i(\widetilde W,\widetilde M')}$$ The lower triangles work similarly. The middle square only commutes up to sign: let~$a\in C_{i+1}(\widetilde w)$ be a relative cycle in~$W$, so $\partial a=b+c\in C_i(\widetilde M)\oplus C_i(\widetilde M')$. Then the upper path maps $[a]$ to $[b]$, and the lower path maps $[a]$ to $[c]=-[b+\partial a]=-[b]$. To repair this, put an additional minus sign on all boundary arrows landing in $H_\bullet(\widetilde M)$. --~~~~ <wikitex>(\widetilde W,\widetilde M)$, and of the triple $(\widetilde W,\widetilde M\cup\widetilde M',\widetilde M)$, and of the analogous sequences with $\widetilde M$ and $\widetilde M'$ swapped. Because $\widetilde M$ and $\widetilde M'$ are disjoint, we have an isomorphism $H_k(\widetilde M\cup\widetilde M',\widetilde M')\cong H_k(\widetilde M)$ and another one with $\widetilde M$ and $\widetilde M'$ swapped. Both are implicit in the braid.

The squares that are half visible on the left and right hand side obviously commute. The upper triangles commute because of the following diagram,

The lower triangles work similarly.

The middle square only commutes up to sign: let~$a\in C_{i+1}(\widetilde w)$ be a relative cycle in~$W$, so $\partial a=b+c\in C_i(\widetilde M)\oplus C_i(\widetilde M')$. Then the upper path maps $[a]$ to $[b]$, and the lower path maps $[a]$ to $[c]=-[b+\partial a]=-[b]$. To repair this, put an additional minus sign on all boundary arrows landing in $H_\bullet(\widetilde M)$.

2) We have learned that $W$ is homotopy equivalent to $M\cup_{g|}D^{k+1}$. Let $\tilde g|\colon S^k\to\widetilde M$ be a lift (ok if $k\ge2$) If we pass to the universal cover of $M$, we are attaching a $\pi$-cell $\pi\times D^{k+1}$ by $(\gamma,t)\mapsto\gamma(\tilde g(t))$. Excision gives

$$\displaystyle H_i(\widetilde W,\widetilde M)\cong H_i(\widetilde M\cup_{\pi\times S^k}(\pi\times D^{k+1}),\widetilde M) \cong \Z\pi H_i(D^{k+1},S^k).$$

Similarly, swapping $W$ is homotopy equivalent to $M'\cup_{S^{n-k}}D^{n+1-k}$. This gives the second assertion.

--~~~~